Question

To find: The centroid \((\bar x,\bar y)\) of the region bounded by the given curve: \(y = 2 - {x^2}\), \(y = x\).

Short answer

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Step-by-step solution

The concept used to find the centroid of the region bounded by curve

The centroid of a plane region is the centre point of the region over the interval [a, b]. In order to calculate the coordinates of the centroid, we'll need to calculate the area of the region first.

Plot the graph for the given equations using the calculation as follows

Show the equations as below:

\(y = 2 - {x^2}\) …… (1)

\(y = x\) …… (2)

Calculate \(y\) value using Equation (1)

Substitute 0 for \(x\) in Equation (1).

\(\begin{array}{}y = 2 - {0^2}\\ = 2\end{array}\)

Hence the co-ordinate of (x, y) is\((0,2)\).

Calculate \(y\) value using Equation (1)

Substitute 1 for \(x\) in Equation (1).

\(\begin{array}{}y = 2 - {1^2}\\ = 1\end{array}\)

Hence the co-ordinate of (x, y) is\((1,1)\).

Calculate \(y\) value using Equation (2)

Substitute 0 for \(x\) in Equation (2).

\(y = 0\)

The co-ordinate of (x, y) is\((0,0)\).

Calculate \(y\) value using Equation (2).

Substitute 1 for \(x\) in Equation (2).

\(y = 1\)

The co-ordinate of (x, y) is\((1,1)\).

Similarly calculate the coordinate values up to bound the region in the graph.

Draw the region as in Figure 1.

Calculate the area of the region bounded by the given curve

…… (3)

Calculate the \(x\) values:

Substitute \(x\) for \(y\) in Equation (1)

\[\begin{array}{}x = 2 - {x^2}\\{x^2} + x - 2 = 0\\{x^2} - x + 2x - 2 = 0\\x(x - 1) + 2(x - 1) = 0\\(x - 1)(x + 2) = 0\\{x_1} = 1\\{x_2} = - 2\end{array}\]

Substitute \(( - 2)\) for a, \(1\) for b, \(2 - {x^2}\) for\([f(x)]\), and \(x\) for \([g(x)]\) in Equation (3).

\(\begin{array}{}A = \int_{ - 2}^1 {\left( {2 - {x^2} - x} \right)} dx\\ = \int_{ - 2}^1 {\left( { - {x^2} - x + 2} \right)} dx\end{array}\) …… (4)

Integrate equation (4)

\(\begin{array}{}A = \left[ { - \frac{{{x^{2 + 1}}}}{{2 + 1}} - \frac{{{x^{1 + 1}}}}{{1 + 1}} + 2x} \right]_{ - 2}^1\\ = \left[ { - \frac{1}{3}{x^3} - \frac{1}{2}{x^2} + 2x} \right]_{ - 2}^1\\ = \left[ { - \frac{1}{3}{{(1)}^3} - \frac{1}{2}{{(1)}^2} + 2(1)} \right] - \left[ { - \frac{1}{3}{{( - 2)}^3} - \frac{1}{2}{{( - 2)}^2} + 2( - 2)} \right]\\ = \left[ {\left( {\frac{{ - 1}}{3} - \frac{1}{2} + 2} \right) - \left( {\frac{8}{3} - 2 - 4} \right)} \right]\\ = \frac{7}{6} - \left( {\frac{{ - 10}}{3}} \right)\\ = \frac{{27}}{6}\\ = \frac{9}{2}\end{array}\)

Calculate the \((\bar x)\) coordinate of centroid

For calculating the \((\bar x)\) coordinate of centroid:

\(\bar x = \frac{1}{A}\int_a^b x [f(x) - g(x)]dx\) …… (5)

Substitute (-2) for a, 1 forb, \(\frac{9}{2}\) for A, \(2 - {x^2}\) for\([f(x)]\), and \(x\) for \([g(x)]\) in Equation (5).

\(\begin{array}{}\bar x = \frac{1}{{\frac{9}{2}}}\int_{ - 2}^1 x \left( {2 - {x^2} - x} \right)dx\\ = \frac{2}{9}\int_{ - 2}^1 x \left( { - {x^2} - x + 2} \right)dx\\ = \frac{2}{9}\int_{ - 2}^1 {\left( { - {x^3} - {x^2} + 2x} \right)} dx\end{array}\) …… (6)

Integrate Equation (6).

\(\bar x = \frac{2}{9}\left[ { - \frac{{{x^{3 + 1}}}}{{3 + 1}} - \frac{{{x^{2 + 1}}}}{{2 + 1}} + \frac{{2{x^{1 + 1}}}}{{1 + 1}}} \right]_{ - 2}^1\)

\( = \frac{2}{9}\left[ { - \frac{1}{4}{x^4} - \frac{1}{3}{x^3} + \frac{2}{2}{x^2}} \right]_0^1\)

\( = \frac{2}{9}\left\{ {\left[ { - \frac{1}{4}{{(1)}^4} - \frac{1}{3}{{(1)}^3} + {{(1)}^2}} \right] - \left[ { - \frac{1}{4}{{( - 2)}^4} - \frac{1}{3}{{( - 2)}^3} + {{( - 2)}^2}} \right]} \right\}\)

\( = \frac{2}{9}\left[ {\left( { - \frac{1}{4} - \frac{1}{3} + 1} \right) - \left( { - \frac{{16}}{4} + \frac{8}{3} + 4} \right)} \right]\)

\( = \frac{2}{9}\left( {\frac{5}{{12}} - \frac{8}{3}} \right)\)

\( = \frac{2}{9}\left( { - \frac{9}{4}} \right)\)

\( = - \frac{1}{2}\)

Calculate the \((\bar y)\) coordinate of centroid

For calculating the \((\bar y)\) coordinate of centroid:

\(\bar y = \frac{1}{A}\int_a^b {\frac{1}{2}} \left\{ {{{[f(x)]}^2} - {{[g(x)]}^2}} \right\}dx\) …… (7)

Substitute \(( - 2)\) for a, 1 for b, \(\frac{9}{2}\) for A, \(2 - {x^2}\) for\([f(x)]\), and \(x\) for \([g(x)]\) in Equation (7).

\(\begin{array}{}\bar y = \frac{1}{{\frac{9}{2}}}\int_{ - 2}^1 {\frac{1}{2}} \left[ {{{\left( {2 - {x^2}} \right)}^2} - {x^2}} \right]dx\\ = \frac{2}{9} \times \frac{1}{2}\int_{ - 2}^1 {\left( {4 - 4{x^2} + {x^4} - {x^2}} \right)} dx\\ = \frac{1}{9}\int_{ - 2}^1 {\left( {{x^4} - 5{x^2} + 4} \right)} dx\end{array}\) …… (8)

Integrate equation (8)

\(\bar y = \frac{1}{9}\left[ {\frac{{{x^{4 + 1}}}}{{4 + 1}} - \frac{{5{x^{2 + 1}}}}{{2 + 1}} + 4x} \right]_{ - 2}^1\)

\( = \frac{1}{9}\left[ {\frac{1}{5}{x^5} - \frac{5}{3}{x^3} + 4x} \right]_{ - 2}^1\)

\( = \frac{1}{9}\left\{ {\left[ {\frac{1}{5}{{(1)}^5} - \frac{5}{3}{{(1)}^3} + 4(1)} \right] - \left[ {\frac{1}{5}{{( - 2)}^5} - \frac{5}{3}{{( - 2)}^3} + 4( - 2)} \right]} \right\}\)

\( = \frac{1}{9}\left[ {\left( {\frac{1}{5} - \frac{5}{3} + 4} \right) - \left( { - \frac{{32}}{5} + \frac{{40}}{3} - 8} \right)} \right]\)

\( = \frac{1}{9}\left( {\frac{{38}}{{15}} - \frac{{ - 16}}{{15}}} \right)\)

\( = \frac{1}{9}\left( {\frac{{18}}{5}} \right)\)

\( = \frac{2}{5}\)

Hence, the centroid \((\bar x,\bar y)\) of the region bounded by the given curve is\(\left( { - \frac{1}{2},\frac{2}{5}} \right)\).